Question

For which of the following reactions, $\Delta H < \Delta E$ ?
(A) ${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$
(B) $A{g_2}O(s) \to 2Ag(s) + \dfrac{1}{2}{O_2}(g)$
(C) $CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g)$
(D) $C(s) + {O_2}(g) \to C{O_2}(g)$

Answer 1

In thermodynamics, $\Delta H$ is called the standard change in enthalpy. It is the change in enthalpy that occurs in a system when there is a transformation of matter in a chemical reaction i.e. all the reactants are converted into products.

Complete Step-by-step Answer:
The work done can be explained as the product of pressure and the change in volume of the given system against which the system expands. The enthalpy change of a given system during a reaction is given as the change in the internal energy i.e. $\Delta E$ and the work done which is given as $P\Delta V$ . The relation is represented as:
$\Delta H = \Delta E + P\Delta V........(i)$
$\Delta H = \Delta E + \Delta {n_g}RT$
Where, $\Delta {n_g} =$ moles of gas
If the volume of the system becomes constant then the equation (i) becomes $\Delta H = \Delta E$ and if the value of $\Delta {n_g} = - ve$ then the relation will become $\Delta H < \Delta E$ which is the required relation for our given question. So, now we will calculate the value of $\Delta {n_g}$ for every reaction given in option and check which reactions hold the relation true.
For the reaction;
${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$
The value of net moles of gas will be $\Delta {n_g} = 2 - 3 = - 1$
Hence, in this case, $\Delta H < \Delta E$ holds true.
For the reaction,
$A{g_2}O(s) \to 2Ag(s) + \dfrac{1}{2}{O_2}(g)$
The value of net moles of gas will be \Delta {n_g} = \dfrac{1}{2} - 0 = {\raise0.7ex\hbox{ 1 } \\!\mathord{\left/ {\vphantom {1 2}}\right.} \\!\lower0.7ex\hbox{ 2 }}
Hence, in this case, $\Delta H < \Delta E$ does not hold true.
For the reaction,
$CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g)$
The value of net moles of gas will be $\Delta {n_g} = 1 - 1 - \dfrac{1}{2} = - \dfrac{1}{2}$
Hence, in this case, $\Delta H < \Delta E$ holds true.
And for the reaction,
$C(s) + {O_2}(g) \to C{O_2}(g)$
The value of net moles of gas will be $\Delta {n_g} = 1 - 1 = 0$
Hence, in this case, $\Delta H < \Delta E$ does not hold true.
Hence we can say that the correct options for which $\Delta H < \Delta E$ holds true are Option (A) and (C).

Note:
$\Delta E$ is called the change in internal energy of the system. It is the sum of all the kinetic and potential energy of the system. Its value is positive if the work is done on the system from the surroundings and it is negative if the work is done by the system.